We know that $\mathbb C$ is an algebraically closed field with characteristic $0$.

It seems that if a proposition that can be expressed in the language of first-order logic is true for an algebraically closed field with characteristic $0$, then it is true for $\mathbb C$ (and for every algebraically closed field with characteristic $0$).

I am looking for interesting results that would be true over $\mathbb C$ but not true over some algebraically closed field with characteristic $0$. I am more interested in "applied" results than results in field theory.

Additional comment: The initial question is vague (but on purpose). Here is what triggered that question in my mind. Sudbery proved the following result: Let $f(z)$ be a polynomial of degree $N$ with complex coefficients, and let $f^{(r)}(z)$ denote the $r$th derivative of $f(z)$. If $f(z)$ has two or more distinct roots, then $\prod_{i=0}^{N}f^{(r)}(z)$ has at least $N+1$ distinct roots. This result does not seem to generalize easily to algebraically closed fields with characteristic $0$.

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    $\begingroup$ Not every algebraically closed field of characteristic zero is in bijection (just in the set theoretical sense, not any kind of algebraic-structure related isomorphism) with $\mathbb C$. Is that a sufficient distinction? (i.e. some of them are "bigger"). $\endgroup$ Feb 16, 2012 at 3:34
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    $\begingroup$ @Nathan: Can you elaborate on your "additional comment"? That theorem (schema) looks like exactly the sort of thing that elementary equivalence is good for. For each $N$, the claim is a statement in the first-order language of a field. Since the claim holds over $\mathbb{C}$ for each $N$, the theorem must hold for any algebraically closed field. $\endgroup$
    – user14972
    Feb 16, 2012 at 4:35
  • $\begingroup$ @Hurkyl, at the end of blms.oxfordjournals.org/content/5/1/13.extract the author makes a conjecture about algebraically closed field. $\endgroup$ Feb 16, 2012 at 4:39

4 Answers 4


If you want statements that are field-theoretic in nature, then a useful contrast is between the algebraically closed field $\overline{\mathbb Q}$ (the algebraic closure of $\mathbb Q$ in $\mathbb C)$ and $\mathbb C$. The former is the smallest algebraically closed field of char. zero, while $\mathbb C$ is huge in comparison.

For example, $\mathbb C$ contains a sequence of elements $x_i$ which are all mutually algebraically independent (in other words, it contains a copy of the field $\mathbb Q(x_1,\ldots,x_n,\ldots)$ of rational functions over $\mathbb Q$ in a countable number of variables), while $\overline{\mathbb Q}$ does not contain even one element that is algebraically independent of $\mathbb Q$.

Added: Note that the above property makes it much easier to prove the Nullstellenstatz over $\mathbb C$ then over $\overline{\mathbb Q}$, for example.


I (too) find the original question quite vague: there are so many differences between $\overline{\mathbb{Q}}$ and $\mathbb{C}$ that I wouldn't know where to begin. What is more interesting is that by model theory the first order theories of these fields agree.

Compare this with the "Lefschetz Principle" in algebraic geometry: if $K_1$ is any algebraically closed field of characteristic zero and infinite absolute transcendence degree, then "algebraic geometry is the same over $K_1$ as it is over $\mathbb{C}$". The statement in quotation marks is not precise but should definitely be construed to mean much more than just equality of the first order theories, as classical algebraic geometers well know (this is the principle behind Weil's universal domains). Model theorists have tried to enunciate the Lefschetz Principle more precisely, although to the best of my knowledge never in a fully definitive way.

With regard to your "Additional comment": for every positive integer $N$ you have written down a first order statement $P(N)$ in the language of fields which Sudbery showed holds true over $\mathbb{C}$ using, in part, convexity arguments. I claim that it follows immediately from the completeness of $\operatorname{ACF}_0$ that for all $N \in \mathbb{Z}^+$, $P(N)$ holds over any algebraically closed field of characteristic zero (even though there is apparently no sensible notion of "convexity" in this context -- i.e., the model theory is actually helping out here!). Further, by Ax's Transfer Principle, for each fixed $N$, there must exist a positive integer $M(N)$ such that for all primes $p \geq M(N)$, the statement $P(N)$ holds over any algebraically closed field of characteristic $p$. You can find (what I hope is) a very detailed, elementary presentation of this transfer principle in $\S 3.5$ of these notes.

Now, let me draw your attention to the following MathSciNet Review:

MR1152642 (92j:11144) Bartocci, Umberto; Vipera, Maria Cristina On the Gauss-Lucas lemma in positive characteristic. (Italian summary) Atti Accad. Naz. Lincei Rend. Cl. Sci. Fis. Mat. Natur. (8) 82 (1988), no. 2, 211–216 (1989).

If $f(x)$ is a nonconstant polynomial with coefficients in the field of complex numbers of degree $n$, then it has at least one root α; for all k in the range μ≤k≤n, where μ is the multiplicity of α, we have f(k)(α)/=0 (such a root is called a "free'' root of f). This is a consequence of the Gauss-Lucas lemma which was proved by A. Sudbery [Bull. London Math. Soc. 5 (1973), 13--17; MR0320288 (47 #8827)]. Sudbery conjectured that this property remains true for polynomials with coefficients in a field of positive characteristic p>n. Bartocci [Atti Accad. Naz. Lincei Rend. Cl. Sci. Fis. Mat. Natur. (8) 56 (1974), no. 6, 856--858; MR0389865 (52 #10695)] proved that this conjecture is false, but true for p>C(n). In this paper it is shown that, if $n<p<2n−2$, then there exist polynomials which do not have free roots at all. (Reviewed by M. Marden)

That is, the part of Sudbery's conjecture that follows immediately from standard model theory was proven (not using model theory, so far as I know) in a paper published one year later. Much later it was shown that the stronger qualitative form of Sudbery's conjecture is very far from being true. (This situation is somewhat similar to what happened with Artin's Conjecture on $p$-adic fields: see $\S 7$ of the same notes.)

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    $\begingroup$ @Nathan: you're welcome. Actually I didn't know about the Gauss-Lucas Theorem until your question prompted me to look up Sudbery's paper....what a nice result! So thanks to you too. $\endgroup$ Feb 16, 2012 at 14:08

Here is an example of a non-first order statement which is true over $\mathbb{C}$ (and in fact, any uncountable algebraically closed field) but not, say, over $\overline{\mathbb{Q}}$.

Let $k$ be an algebraically closed field. Recall Schur's lemma: if $A$ is a finite-dimensional $k$-algebra, and $V$ an irreducible $A$-module, then $\hom_A(V, V) = k$ consists of scalars. Under certain circumstances, it is possible to extend this to infinite-dimensional algebras. Dixmier's lemma states that if $k$ is uncountable, then if $A$ is a countably dimensional irreducible $A$-module, we have $\hom_A(V, V) = k$. The usual proof of Schur's lemma (which relies on finding an eigenvalue of an endomorphism) breaks down, but we can argue as follows: $\hom_A(V, V)$ is a division ring, and if it contains an element $T \notin k$, there is a (commutative) field $k(T) \subset \hom_A(V, V)$. But $\hom_A(V, V)$ is countably dimensional (since it injects into $\hom_k(v, V)$ for any $v \neq 0$ in $V$, by irreducibility), whereas $k(T)$ is not countably dimensional if $k$ is uncountable. The partial fractions $1/(T - \lambda), \lambda \in k$ are linearly independent.

The essential idea of the proof is that a proper extension of $k$ must be uncountably dimensional. Note that the restriction on $k$ is necessary: otherwise, for instance, we could take the countably-dimensional $\overline{\mathbb{Q}}$-algebra $\overline{\mathbb{Q}}(t)$ and consider it as a simple module over itself. Then $\hom_\overline{\mathbb{Q}}(t)(\overline{\mathbb{Q}}(t), \overline{\mathbb{Q}}(t)) = \overline{\mathbb{Q}}(t)$. So Dixmier's lemma is false without the cardinality hypotheses.

As Matt E observes, the Nullstellensatz is much easier to prove over uncountable fields: the key lemma in the Nullstellensatz, in fact, is the statement that a finitely generated algebra over a field which is itself a field is an algebraic extension. This cardinality argument gives a quick proof of that.

  • $\begingroup$ I think this proof of the Nullstellensatz for uncountable algebraically closed fields is certainly interesting (and when I get around to putting a chapter on model theory in my commutative algebra notes, I will probably want to include it). But do you really think it's much easier than other proofs of Zariski's Lemma? What about the one using the Artin-Tate Lemma? Certainly that proof is quicker than yours if you develop the relevant model-theoretic transfer arguments as part of your proof. $\endgroup$ Feb 16, 2012 at 5:11
  • $\begingroup$ Dear Akhil, Very nice! Cheers, $\endgroup$
    – Matt E
    Feb 16, 2012 at 5:19
  • $\begingroup$ Very nice indeed Akhil. $\endgroup$ Feb 16, 2012 at 12:50
  • $\begingroup$ @Matt: Dear Matt, Thanks! By the way, I'm sorry I missed your talk at Harvard a few days back. Had I paid more attention to the list of seminars I would have certainly been there. $\endgroup$ Feb 16, 2012 at 13:05
  • $\begingroup$ @Pete: I was unaware of the argument via the Artin-Tate lemma; thanks for bringing it to my attention! If I ever get around to reviving the "CRing project" I will be sure to include this somewhere. As for "easy": that's of of course subjective, but as someone who has difficulty remembering purely algebraic arguments well, I find occasional "tricks" to be useful, especially when they connect to a deeper story. I'm not very familiar with model theory but this kind of argument is quite suggestive (e.g. reducing from char. 0 to $\mathbb{C}$ in alg. geo., where analytic techniques can be used). $\endgroup$ Feb 16, 2012 at 13:22

As stated this question is incredibly vague, and answering is difficult. Regardless, here are a few answers:

  • $\mathbb{C}$ is complete while $\overline{\mathbb{Q}}$ is not.
  • $\mathbb{C}$ is uncountable while $\overline{\mathbb{Q}}$ is countable.
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    $\begingroup$ The field ${\mathbf C}_p$ admits an archimedean absolute value since ${\mathbf C}$ and ${\mathbf C}_p$ are isomorphic fields. As valued fields, with their usual absolute values, they are not isomorphic, but purely as fields they are. $\endgroup$
    – KCd
    Feb 16, 2012 at 3:40
  • $\begingroup$ @KCd: Ah, of course. Thanks. $\endgroup$ Feb 16, 2012 at 3:55

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